+0  
 
+1
128
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avatar+142 

 

Let $a,$ $b,$ and $c$ be positive real numbers such that $a > b$ and $a + b + c = 4.$

 

Find the minimum value of $4a + 3b + \frac{c^3}{(a - b)b}.$

 

How do I find the minimum value of this? Any help I will appreciate, and I will get back to you extremely quickly. Thank you!

 May 12, 2021
 #1
avatar+759 
+3

AM-GM can likely help you here lol

 

the equation would go $a + \dfrac{c^3}{(a - b)b} \geq 3 \sqrt[3]{ \dfrac{(a - b)bc^3}{(a - b)b}} = 3c.$

this'd imply that, $4a + 3b + \frac{c^3}{(a - b)b} \geq 3a + 3b + 3c$

 

the rhs is 3(a + b + c), or 12 - thus, 12 is the local minimum.

 May 12, 2021
 #2
avatar+142 
+1

THANK you!!!

apologies for the long time for reply... i was on vacation. I will be sure to study up on my artihmetic-geometric mean inequalities  😅

HighSchoolDx  May 16, 2021

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